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Introduction to Le Chatelier's Principle

For the react...

Question

For the reaction,
$C_{3} H_{8 (g)} + 5 O_{2 (g)} ⇌ 3 C O_{2 (g)} + 4 H_{2} O_{(l i q)}$ at constant temperature, $Δ H - Δ E$ is:

+ RT

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- 3 RT

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+ 3 RT

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- RT

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Solution

The correct option is B $- 3 RT$
For the reaction,
$C_{3} H_{8 (g)} + 5 O_{2 (g)} ⇌ 3 C O_{2 (g)} + 4 H_{2} O_{(l i q)}$
$Δ n_{(g)} = 3 - 6 = - 3$
As we know,
$Δ H = Δ E + P Δ V$
$\Rightarrow Δ H - Δ E = P Δ V$
$\Rightarrow Δ H - Δ E = Δ n_{(g)} R T = - 3 R T$

Suggest Corrections

Similar questions

C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 C O_{2} (g) + 4 H_{2} O (l)

Δ H - Δ E

C_{3} H_{8} (g) + 5 O_{2} (g)

⟶

3 C O_{2} (g) + 4 H O (I)

Δ

H -

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C_{3} H_{8} (g) + 5 O_{2} (g) \to 3 {C O}_{2} (g) + 4 H_{2} O (l)

Δ H - Δ U

Complete and balance the given reaction:

$C_{3} H_{8} (g) + O_{2} (g) \to$

Δ H

Δ U

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